3.0unit3-acid_and_bases
TRANSCRIPT
Chemistry 2
Unit 3 Unit 3 Acids-Bases and SaltAcids-Bases and Salt
Pajuzi AwangPajuzi AwangIPG KAMPUS DATO’ RAZALI ISMAILIPG KAMPUS DATO’ RAZALI ISMAIL
Acids and BasesAcids and Bases
Bronsted Lowry Acids and BasesBronsted Lowry Acids and Bases Autoionization of WaterAutoionization of Water pHpH Strong Acids and BasesStrong Acids and Bases Weak Acids and BasesWeak Acids and Bases Ionization ConstantsIonization Constants Lewis Acids and BasesLewis Acids and Bases The Common Ion EffectThe Common Ion Effect BuffersBuffers TitrationsTitrations
ReviewReview
Arrhenius AcidArrhenius Acid
– Substance that increases the Substance that increases the concentration of Hconcentration of H++ ions when ions when dissolved in waterdissolved in water
HCl (g) HHCl (g) H++ (aq) + Cl (aq) + Cl-- (aq) (aq)H2O
ReviewReview
Arrhenius BaseArrhenius Base
– Substance that increases the Substance that increases the concentration of OHconcentration of OH-- ions when ions when dissolved in water.dissolved in water.
NaOH (s) NaNaOH (s) Na++ (aq) + OH (aq) + OH-- (aq) (aq)H2O
ReviewReview
The Arrhenius definition of acids and The Arrhenius definition of acids and bases has limitations:bases has limitations:– Limited to aqueous solutionsLimited to aqueous solutions
Two other common definitions for acids Two other common definitions for acids and bases.and bases.– Bronsted-Lowry acids and basesBronsted-Lowry acids and bases– Lewis acids and bases (pg 125)Lewis acids and bases (pg 125)
Bronsted-Lowry Acids & Bases Bronsted-Lowry Acids & Bases (pg 96)(pg 96) Bronsted-Lowry AcidBronsted-Lowry Acid– A substance (ion or molecule) that can A substance (ion or molecule) that can
transfer a proton (Htransfer a proton (H++ ion) to another ion) to another substancesubstance
HCl (g) + HHCl (g) + H22OO (l) H(l) H33OO++ (aq) + Cl (aq) + Cl-- (aq) (aq)
B-L acid
Bronsted-Lowry Acids & Bases Bronsted-Lowry Acids & Bases (pg 96)(pg 96) Bronsted-Lowry BaseBronsted-Lowry Base– A substance (ion or molecule) that can A substance (ion or molecule) that can
accept a proton (Haccept a proton (H++ ion) from another ion) from another substancesubstance
HCl (g) + HHCl (g) + H22OO (l) H(l) H33OO++ (aq) + Cl (aq) + Cl-- (aq) (aq)
B-L acid B-L base
Notice that H2O acts as a base, accepting a proton from HCl.
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Bronsted-Lowry acid:Bronsted-Lowry acid:– Molecule or ionMolecule or ion– Must have H atom in the formula that Must have H atom in the formula that
can be lost as Hcan be lost as H++ ion ion
Bronsted-Lowry Base:Bronsted-Lowry Base:– Molecule or ionMolecule or ion– Must have lone pair of electrons Must have lone pair of electrons
(nonbonding pair) that can bind a H(nonbonding pair) that can bind a H++ ionion
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
HH++ ion in Water ion in Water
+ e-
H atomH atom HH++ ion ion
+
Lose e- Proton with no surrounding valence electrons
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
HH+ + ion interacts strongly with the ion interacts strongly with the nonbonding pairs of electrons on water nonbonding pairs of electrons on water moleculesmolecules– Forms hydrated hydrogen ionForms hydrated hydrogen ion
““hydronium ionhydronium ion””
HH+ + + O – H + O – H H – O – H H – O – H ++
HH HH
Hydronium ion
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
The hydrated proton (hydronium ion) is The hydrated proton (hydronium ion) is responsible for the characteristic responsible for the characteristic properties of aqueous solutions of acids.properties of aqueous solutions of acids.
HH33OO+ + is a more realistic depiction of the is a more realistic depiction of the
hydrogen ion in solutionhydrogen ion in solution– for convenience we often use Hfor convenience we often use H++ (aq) (aq)
to depict the hydrated hydrogen ionto depict the hydrated hydrogen ion
HCl (g) + H20 (l) H3O+ (aq) + Cl- (aq)
HCl (aq) H+ (aq) + Cl- (aq)
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases Bronsted-Lowry theory is applicable to Bronsted-Lowry theory is applicable to
reactions that occur in both aqueous and reactions that occur in both aqueous and non-aqueous solution.non-aqueous solution.
HH H H ++
H – N + H – Cl H – N + H – Cl H – N – H + Cl H – N – H + Cl--
HH HH
HCl (aq) + NH3 (aq) NH4+ (aq) + Cl- (aq)
HCl (g) + NH3 (g) NH4Cl (s)
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Consider the following reactions:Consider the following reactions:
HH22O (l) + HCl (g) O (l) + HCl (g) H H33OO++ (aq) + Cl(aq) + Cl-- (aq) (aq)
NHNH33 (g) + H (g) + H22O (l) O (l) NH NH44++ (aq) + OH (aq) + OH-- (aq) (aq)
acid
base
In the first reaction, HIn the first reaction, H22O acts as a O acts as a base.base.
In the second reaction, HIn the second reaction, H22O acts as an O acts as an
acid.acid.
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Water is Water is amphotericamphoteric::– Capable of acting as either an acid or a Capable of acting as either an acid or a
base.base.
In any acid-base equilibrium, both In any acid-base equilibrium, both forward and reverse reactions involve forward and reverse reactions involve proton transfers.proton transfers.
HNOHNO22 (aq)(aq) + H + H22OO (l)(l) NO NO2 2 –– (aq)(aq) + H + H33OO
++ (aq)(aq)
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
In any acid-base equilibrium, both In any acid-base equilibrium, both forward and reverse reactions involve forward and reverse reactions involve proton transfers.proton transfers.
HNOHNO22 (aq)(aq) + H + H22OO (l)(l) NO NO2 2 –– (aq)(aq) + H + H33OO
++ (aq)(aq)
Forward Reaction:Forward Reaction:– B-L acid =B-L acid =– B-L base =B-L base =
Reverse Reaction:Reverse Reaction:– B-L acid =B-L acid =– B-L base =B-L base =
HNOHNO22
HH22OO
HH33OO++
NONO22--
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
HNOHNO22 (aq)(aq) + H + H22OO (l)(l) NO NO2 2 –– (aq)(aq) + H + H33OO
++ (aq)(aq)
baseacid acidbase
When HNOWhen HNO22 acts as an acid, it loses a acts as an acid, it loses a
proton and forms NOproton and forms NO22 – –
– NONO22-- acts as a base acts as a base
When HWhen H22O acts as a base, it gains a proton O acts as a base, it gains a proton
and forms Hand forms H33OO++
– HH33OO++ acts as an acid acts as an acid
Bronsted-Lowry Acids & Bases Bronsted-Lowry Acids & Bases (pg 97)(pg 97)HNOHNO22 (aq)(aq) + H + H22OO (l)(l) NO NO2 2
–– (aq)(aq) + H + H33OO++ (aq)(aq)
baseacid acidbase
HNOHNO22 and NO and NO22-- are called a are called a conjugate conjugate
acid-base pair.acid-base pair.
HH22O and HO and H33OO++ are also a are also a conjugate acid-conjugate acid-
base pair.base pair.
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases Conjugate acid-base pair:Conjugate acid-base pair:– An acid and a base that differ only in An acid and a base that differ only in
the presence or absence of a single the presence or absence of a single protonproton HNOHNO22 and NO and NO22
--
HH33OO++ and H and H22OO HCOHCO33
-- and COand CO33 2-2-
NHNH44++ and NH and NH33
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Every acid has a Every acid has a conjugate base:conjugate base:– The base formed by removing a proton The base formed by removing a proton
from an acidfrom an acid Conjugate base of HConjugate base of H22OO–OH OH --
Conjugate base of HConjugate base of H22SOSO44
–HSOHSO44 --
Conjugate base of HClOConjugate base of HClO44
–ClOClO4 4 ––
NOTE: The conjugate base is always NOTE: The conjugate base is always shown on the product side.shown on the product side.
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Every base has a Every base has a conjugate acid:conjugate acid:– The acid formed when a base gains a The acid formed when a base gains a
protonproton Conjugate acid of HConjugate acid of H22OO
–HH33O O ++
Conjugate acid of SOConjugate acid of SO4 4 2-2-
–HSOHSO44 --
Conjugate acid of CN Conjugate acid of CN --
–HCNHCN NOTE: The conjugate acid is always NOTE: The conjugate acid is always
shown on the product side.shown on the product side.
Chemistry 2
Unit 3 Unit 3 Acids-Bases and SaltAcids-Bases and Salt
Pajuzi AwangPajuzi AwangIPG KDRI-Lecture 2IPG KDRI-Lecture 2
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
HNOHNO22 (aq)(aq) + + HH22OO (l)(l) NONO2 2 –– (aq)(aq) + + HH33OO
++ (aq)(aq)
baseacid conjugatebase
Conjugateacid
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Example:Example: What is the conjugate base of: What is the conjugate base of:
HClOHClO44
HH33POPO44
HSOHSO44 --
ClO4 -
H2PO4 -
SO4 2-
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Example:Example: What is the conjugate acid of: What is the conjugate acid of:
HSOHSO44 - -
CNCN - -
OH OH --
H2SO4
HCN
H2O
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Example:Example: Identify the acid and base on the Identify the acid and base on the reactant side and the conjugate acid and reactant side and the conjugate acid and conjugate base on the product side of the conjugate base on the product side of the following reaction.following reaction.
HSOHSO44–– (aq) + CO (aq) + CO33
2-2- (aq) SO (aq) SO442-2- (aq) + HCO (aq) + HCO33
2-2- (aq) (aq)
baseacid Conjugateacid
Conjugatebase
In any acid-base equilibrium, both In any acid-base equilibrium, both forward and reverse reactions involve forward and reverse reactions involve proton transfers.proton transfers.
HNOHNO22 (aq)(aq) + H + H22OO (l)(l) NO NO2 2 –– (aq)(aq) + H + H33OO
++ (aq)(aq)
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
In any acid-base equilibrium, both the In any acid-base equilibrium, both the forward and reverse reactions involve forward and reverse reactions involve proton transfers.proton transfers.
HNOHNO22 (aq)(aq) + H + H22OO (l)(l) NO NO2 2 –– (aq)(aq) + H + H33OO
++ (aq)(aq)
How can we predict the position of the How can we predict the position of the chemical equilibrium?chemical equilibrium?
baseacid Conjugateacid
Conjugatebase
Bronsted Lowry Acids & BasesBronsted Lowry Acids & Bases
The relative strengths of the acid and the The relative strengths of the acid and the conjugate acid can be used to predict the conjugate acid can be used to predict the position of the equilibrium.position of the equilibrium.
Equilibrium favors the formation of the Equilibrium favors the formation of the weaker acidweaker acid
– The stronger acid more effectively The stronger acid more effectively loses a proton than the conjugate acidloses a proton than the conjugate acid
ACID-BASE PROPERTIES ACID-BASE PROPERTIES (pg 98)(pg 98)
Autoionization and the Ion-Product of Autoionization and the Ion-Product of WaterWater The pH and pOH ScaleThe pH and pOH Scale
Autoionization of WaterAutoionization of Water
One of the most important properties of One of the most important properties of water is its ability to act as either an acid water is its ability to act as either an acid or a baseor a base
In the presence of an acid, water acts as In the presence of an acid, water acts as a base by accepting a Ha base by accepting a H++
HCl (g) + HHCl (g) + H22OO H H33OO++ (aq) + Cl (aq) + Cl - - (aq) (aq)
In the presence of a base, water acts as In the presence of a base, water acts as an acid, donating a Han acid, donating a H++
NHNH33 (g) + H (g) + H22OO NHNH44++ (aq) + OH (aq) + OH - - (aq) (aq)
Autoionization of WaterAutoionization of Water
Water molecules can also donate a proton to Water molecules can also donate a proton to another water molecule in a process called another water molecule in a process called autoionization.autoionization.
H O + H O H O H H O + H O H O H ++ + OH + OH --
HH HH HH
Autoinization
Autoionization of WaterAutoionization of Water
Autoionization:Autoionization:– the process in which water the process in which water
spontaneously forms low spontaneously forms low concentrations of Hconcentrations of H++ and OH and OH -- ions by ions by proton transfer from one water proton transfer from one water molecule to anothermolecule to another
2 H2O (l) H3O+ (aq) + OH- (aq)
Autoionization of WaterAutoionization of Water
Autoionization of water is a very rapid Autoionization of water is a very rapid processprocess– both forward and reverse reactions both forward and reverse reactions
occur rapidlyoccur rapidly
At any given time only a very small At any given time only a very small number of water molecules are ionizednumber of water molecules are ionized
If every letter in our text represented a water molecule, you would have to look through about 50 texts to find one H3O+!
Autoionization of WaterAutoionization of Water
An equilibrium constant expression can An equilibrium constant expression can be written for the autoionization of water:be written for the autoionization of water:
2 H2 H22O (l) HO (l) H33OO++ (aq) + OH (aq) + OH - - (aq) (aq)
KKww = [H = [H33OO++] [OH] [OH - -] = [H] = [H+ + ] [OH] [OH- - ]]
where Kwhere Kww = ionization constant for water = ionization constant for water
= ion-product constant= ion-product constant
Note: H2O is excluded because it is a pure liquid.
Autoionization of WaterAutoionization of Water
The ionization constant expression for The ionization constant expression for water (and its value) is useful not only for water (and its value) is useful not only for pure water but also for pure water but also for all dilute aqueous all dilute aqueous solutions.solutions.– Use it to calculate the [HUse it to calculate the [H+ + ] and [OH] and [OH-- ] ]
for dilute solutions.for dilute solutions.
For all dilute aqueous solutions:For all dilute aqueous solutions:
KKww = [H = [H+ + ] [OH] [OH-- ] = 1.0 x 10 ] = 1.0 x 10-14-14 at 25at 25ooCC
Autoionization of WaterAutoionization of Water
In a neutral solution:In a neutral solution:– [H[H+ + ] = [OH] = [OH-- ] ]
In an acidic solution:In an acidic solution:– [H[H+ + ] > [OH] > [OH-- ] ]
In a basic solutionIn a basic solution– [H[H+ + ] < [OH] < [OH-- ] ]
Autoionization of WaterAutoionization of Water
Example:Example: Calculate the [HCalculate the [H+ + ] and [OH] and [OH-- ] ] in a neutral in a neutral solution at 25solution at 25ooC.C.
[H[H+ + ] [OH] [OH-- ] = (x) (x) = 1.0 x 10 ] = (x) (x) = 1.0 x 10-14-14
xx22 = 1.0 x 10 = 1.0 x 10-14-14
x = 1.0 x 10x = 1.0 x 10-7 -7 MM
In a neutral solution: [H[H+ + ] = [OH] = [OH-- ] = 1.0 x 10 ] = 1.0 x 10-7 -7 MM
Chemistry 2
Unit 3 Unit 3 Acids-Bases and SaltAcids-Bases and Salt
Pajuzi AwangPajuzi AwangIPG KDRI-Lecture 3IPG KDRI-Lecture 3
Autoionization of WaterAutoionization of Water
In a neutral solution:In a neutral solution:– [H[H+ + ] = [OH] = [OH-- ] = 1.0 x 10 ] = 1.0 x 10-7-7 M at 25 M at 25ooCC
In an acidic solution:In an acidic solution:– [H[H+ + ] > [OH] > [OH-- ] ]– [H[H+ + ] > 1.0 x 10] > 1.0 x 10-7-7 M at 25 M at 25ooCC
In a basic solutionIn a basic solution– [H[H+ + ] < [OH] < [OH-- ] ]– [H[H+ + ] < 1.0 x 10] < 1.0 x 10-7-7 M at 25 M at 25ooCC
Autoionization of WaterAutoionization of Water
Example: Example: What is the [HWhat is the [H+ + ] at 25] at 25ooC for a C for a solution in which [OHsolution in which [OH-- ] = 0.010 M. ] = 0.010 M.
[[H+ ] [OH- ] = 1.0 x 10-14 at 25oC
Given: Given: [OH[OH--] = 0.010 M] = 0.010 M
Find:Find: [H [H+ + ] ]
[H[H+ + ] (0.010) = 1.0 x 10] (0.010) = 1.0 x 10-14-14
[H[H++ ] = ] = 1.0 x 101.0 x 10-14-14 = 1.0 x 10 = 1.0 x 10-12-12 M M0.0100.010
Autoionization of WaterAutoionization of Water
Example: Example: What is the [OHWhat is the [OH-- ] at 25 ] at 25ooC in a C in a solution in which [Hsolution in which [H+ + ] = 2.5 x 10] = 2.5 x 10-6-6 M. M.
[H[H+ + ][OH][OH-- ] = 1.0 x 10 ] = 1.0 x 10-14-14
Given: Given: [H[H+ + ] = 2.5 x 10] = 2.5 x 10-6-6 M M
Find:Find: [OH [OH-- ] ]
2.5 x 102.5 x 10-6-6 [OH [OH-- ] = 1.0 x 10 ] = 1.0 x 10-14-14
[OH[OH-- ] = ] = 1.0 x 101.0 x 10-14-14 = 4.0 x 10 = 4.0 x 10-9-9 M M 2.5 x 102.5 x 10-6-6
Chemistry 2
Unit 3 Unit 3 Acids-Bases and SaltAcids-Bases and Salt
Pajuzi AwangPajuzi AwangIPG KDRI-Lecture 3IPG KDRI-Lecture 3
pHpH
Since the [HSince the [H++] is usually very small in ] is usually very small in aqueous solutions, we normally express aqueous solutions, we normally express the [Hthe [H++] in terms of] in terms of pH pH..
pH = - logpH = - log1010 [H [H++]]
pHpH
Example: Example: Calculate the pH of a neutral Calculate the pH of a neutral solution at 25solution at 25ooC.C.
Given: Given: neutral solutionneutral solution
Find:Find: pH pH
Neutral solution:Neutral solution:[H[H++] = 1.0 x 10] = 1.0 x 10-7-7 M M at 25at 25ooCC
pH = - log [HpH = - log [H++]]
pH = - log (1.0 x 10pH = - log (1.0 x 10-7-7) = - (-7.0) = 7.0) = - (-7.0) = 7.0
pHpH
Example: Example: Calculate the pH of a solution Calculate the pH of a solution with [Hwith [H++] = 2.5 x 10] = 2.5 x 10-5-5..
Given: Given: [H[H++] = 2.5 x 10] = 2.5 x 10-5-5
Find:Find: pH pH
pH = - log [HpH = - log [H++]]
pH = - log (2.5 x 10pH = - log (2.5 x 10-5-5) = - (-4.60) = 4.60) = - (-4.60) = 4.60
pHpH
Example: Example: Calculate the pH of a solution Calculate the pH of a solution with [OHwith [OH--] = 6.5 x 10] = 6.5 x 10-5-5..
Given: Given: [OH[OH--] = 6.5 x 10] = 6.5 x 10-5-5
Find:Find: pH pH
[H[H++] [OH] [OH--] = 1.0 x 10] = 1.0 x 10-14 -14 at 25at 25ooCC
[H[H++] = ] = 1.0 x 101.0 x 10-14-14 = 1.54 x 10 = 1.54 x 10-10 -10 MM6.5 x 106.5 x 10-5-5
pH = - log (1.54 x 10pH = - log (1.54 x 10-10-10) = - (-9.8) = 9.8) = - (-9.8) = 9.8
Find [HFind [H++] first.] first.
pH and pOHpH and pOH
The potential of the hydrogen ion was The potential of the hydrogen ion was defined in 1909 as defined in 1909 as the negative of the the negative of the logarithm of logarithm of [H[H++].].
pH = -log[H3O+] pOH = -log[OH-]
-logKW = -log[H3O+]-log[OH-]= -log(1.0x10-14)
KW = [H3O+][OH-]= 1.0x10-14
pKW = pH + pOH= -(-14)
pKW = pH + pOH = 14
pHpH
The negative log is also used to express The negative log is also used to express the magnitude of other small quantities:the magnitude of other small quantities:
pOH = - log [OHpOH = - log [OH- - ]]
pH and pOH are related by the following pH and pOH are related by the following equation that is derived by taking the equation that is derived by taking the negative log of the expression for Knegative log of the expression for Kww
pH + pOH = 14.00pH + pOH = 14.00
pHpH
Example:Example: Calculate the pOH of a solution Calculate the pOH of a solution with [OH with [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M. M.
pOH = - log [OHpOH = - log [OH-- ] ]
pOH = - log (2.5 x 10pOH = - log (2.5 x 10-3-3))
pOH = 2.60pOH = 2.60
pHpH
Example:Example: Calculate the pH of a solution Calculate the pH of a solution with [OH with [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M. M.
Given:Given: [OH [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M M
Find:Find: pHpH
There are two There are two approaches that can approaches that can be used to solve this be used to solve this problem.problem.
Find pOH, thenFind pOH, thenpH + pOH = 14.00pH + pOH = 14.00
11
Find [HFind [H++], then ], then find pH.find pH.22
pHpH
pOH = - log [OHpOH = - log [OH-- ] ]
pOH = - log (2.5 x 10pOH = - log (2.5 x 10-3-3))
pOH = 2.60pOH = 2.60
pH = 14.00 – pOH = 14.00 – 2.60 = 11.40pH = 14.00 – pOH = 14.00 – 2.60 = 11.40
Given:Given: [OH [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M MFind:Find: pHpH
Find pOH, thenFind pOH, then
pH + pOH = 14.00pH + pOH = 14.00
Approach # 1:Approach # 1:
pHpH
[H[H++] [OH] [OH-- ] = 1.00 x 10 ] = 1.00 x 10-14-14
[H[H++] = ] = 1.00 x 101.00 x 10-14-14 = 4.00 x 10 = 4.00 x 10-12-12 M M2.5 x 102.5 x 10-3-3
pH = - log (4.00 x 10pH = - log (4.00 x 10-12-12) = 11.40) = 11.40
Given:Given: [OH [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M MFind:Find: pHpH
Find [H+], thenFind [H+], then
pH = -log [HpH = -log [H++]]
Approach # 2:Approach # 2:
pHpH
Given the pH of a solution, you can also Given the pH of a solution, you can also find the [Hfind the [H++] and the [OH] and the [OH--].].
Since pH = - log [HSince pH = - log [H++], ],
[H[H++] = 10 ] = 10 –pH–pH
Since pOH = - log [OHSince pOH = - log [OH--],],
[OH[OH--] = 10 ] = 10 -pOH-pOH
pHpH
Example:Example: What are the [H What are the [H++] and [OH] and [OH--] for a ] for a solution with a pH of 2.50 at 25solution with a pH of 2.50 at 25ooC?C?
pH = -log [HpH = -log [H++]]
[H[H++] = 10] = 10-pH-pH = 10 = 10 -2.5 -2.5 = 3.16 x 10 = 3.16 x 10-3-3 M M
[H[H++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14
(3.16 x 10(3.16 x 10-3-3 M)[OH M)[OH--] = 1.00 x 10] = 1.00 x 10-14-14
[OH[OH--] = 3.16 x 10] = 3.16 x 10-12-12 M M
Chemistry 2
Unit 3-Next week (19.4 Unit 3-Next week (19.4 STRENGHT OF ACID AND BASE STRENGHT OF ACID AND BASE IN WATER IN WATER
Pajuzi AwangPajuzi AwangIPG KDRI-Lecture 4IPG KDRI-Lecture 4
Bronsted-Lowry Acids & Bases Bronsted-Lowry Acids & Bases (pg 103)(pg 103)
Strong Acid:Strong Acid:– acid that ionizes completelyacid that ionizes completely
no undissociated molecules in no undissociated molecules in solutionsolution
– Conjugate base formed by a strong Conjugate base formed by a strong acid is a very weak baseacid is a very weak base has negligible tendency to gain a has negligible tendency to gain a
protonproton– HCl, HHCl, H22SOSO44
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Weak Acid:Weak Acid:– acid that ionizes partiallyacid that ionizes partially
aqueous solutions contain a mixture of aqueous solutions contain a mixture of acid molecules and the component acid molecules and the component ionsions
– Conjugate base formed by a weak acid is Conjugate base formed by a weak acid is a weak base.a weak base. slight tendency to gain protonsslight tendency to gain protons
– Acetic acid, citric acid, phosphoric acidAcetic acid, citric acid, phosphoric acid
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Substances with negligible aciditySubstances with negligible acidity– substances that contain hydrogen but substances that contain hydrogen but
do not exhibit acidic behaviour in do not exhibit acidic behaviour in waterwater
– conjugate bases are very strong basesconjugate bases are very strong bases react completely with water to form react completely with water to form
OHOH – – ionion
– CHCH44, H, H22, OH, OH--
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
Summary:Summary:– An inverse relationship exists between An inverse relationship exists between
the strength of an acid and its the strength of an acid and its conjugate base or between a base and conjugate base or between a base and its conjugate acid.its conjugate acid.
– The stronger the acid, the weaker the The stronger the acid, the weaker the conjugate baseconjugate base
– The stronger the base, the weaker the The stronger the base, the weaker the conjugate acidconjugate acid
AcidAcidHClHCl
HH22SOSO44
HNOHNO33
HH33OO++ (aq) (aq)
HSOHSO44 - -
NHNH44++
HH22OO
OHOH--
CHCH44
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
BaseBaseClCl --
HSOHSO44 --
NONO33 --
HH22OO
SOSO44 2-2-
NHNH33
OH OH --
OO 2-2-
CHCH33 - -
Weak bases
Strongbases
negligible
Weak acids
negligible
Strongacids
The seven most common strong acids:The seven most common strong acids:– HClHCl hydrochloric acidhydrochloric acid– HBrHBr hydrobromic acidhydrobromic acid– HIHI hydroiodic acidhydroiodic acid– HNOHNO33 nitric acidnitric acid– HClOHClO33 chloric acidchloric acid– HClOHClO44 perchloric acidperchloric acid– HH22SOSO44 sulfuric acidsulfuric acid
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
You must know these acids by name and formula.
Example:Example: Use Fig. 16.4 to predict whether Use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left the equilibrium lies predominantly to the left or to the right:or to the right:
Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases
HSOHSO44-- (aq)(aq) + CO + CO33
22-- (aq)(aq) SO SO442-2- (aq)(aq) + HCO + HCO33
-- (aq)(aq)
acid base Conj.base
Conj.acid
HSO4 - is a stronger acid than HCO3
-.Equilibrium favors formation of the weaker acid.
Equilibrium lies to the right.
Strong AcidsStrong Acids
Strong acid:Strong acid:– Strong electrolyteStrong electrolyte– Ionizes completely in aqueous solutionIonizes completely in aqueous solution
HNOHNO33 (aq)(aq) H H ++ (aq)(aq) + NO + NO33 –– (aq)(aq)
The only significant source of HThe only significant source of H++ ion in ion in an aqueous solution of a strong acid is an aqueous solution of a strong acid is usually the strong acid.usually the strong acid.
HNOHNO33 (aq)(aq) H H ++ (aq)(aq) + NO + NO33 –– (aq)(aq)
In a 0.05 M HNOIn a 0.05 M HNO33 (aq) solution, (aq) solution,
[H[H++] = ] = 0.05 mol HNO0.05 mol HNO33 x x 1 mol H1 mol H++ = 0.05 M= 0.05 M
LL 1 mol HNO1 mol HNO33
Strong AcidsStrong Acids
Consequently, the [HConsequently, the [H++] in a solution of a ] in a solution of a strong monoprotic acid can be strong monoprotic acid can be determined easily using the determined easily using the concentration of the strong acid itself.concentration of the strong acid itself.
Strong AcidsStrong Acids
Example:Example: What is the pH of a 0.25 M HCl What is the pH of a 0.25 M HCl (aq) solution?(aq) solution?
Given:Given: [HCl] = 0.25 M [HCl] = 0.25 M
Find:Find: pHpH
Find [HFind [H++], ], then pHthen pH
[H[H++] = ] = 0.25 mol HCl0.25 mol HCl x x 1 mol H1 mol H++ = 0.25 M = 0.25 MLL 1 mol HCl1 mol HCl
pH = - log (0.25) = 0.60pH = - log (0.25) = 0.60
Strong BasesStrong Bases
Strong BaseStrong Base– strong electrolytestrong electrolyte– ionizes completely in aqueous solutionionizes completely in aqueous solution
NaOH NaOH (aq)(aq) Na Na++ (aq)(aq) + OH + OH-- (aq)(aq)
Common strong basesCommon strong bases– alkali metal hydroxidesalkali metal hydroxides– hydroxides of Ca, Sr, and Bahydroxides of Ca, Sr, and Ba
Strong Bases Strong Bases
Strongly basic solutions are also formed Strongly basic solutions are also formed when certain basic materials react with when certain basic materials react with water to form the OHwater to form the OH-- ion. ion.
OO2-2- (aq)(aq) + H + H22O O (l)(l) 2 OH 2 OH-- (aq)(aq)
HH-- (aq)(aq) + H + H22O O (l) (l) H H22 (g)(g) + OH + OH-- (aq)(aq)
Strong Bases Strong Bases
The pH of an aqueous solution of a The pH of an aqueous solution of a strong base can be determined using the strong base can be determined using the concentration of the strong baseconcentration of the strong base
NaOH NaOH (aq)(aq) Na Na++ (aq)(aq) + OH + OH-- (aq)(aq)
A 0.25 M solution of NaOH has an [OHA 0.25 M solution of NaOH has an [OH--] ] of 0.25 M:of 0.25 M:
0.25 mol NaOH0.25 mol NaOH x x 1 mol OH1 mol OH- - = 0.25 M= 0.25 MLL 1 mol NaOH1 mol NaOH
0.25 M0.25 M
Strong Bases Strong Bases
The pH of the base solution can then be The pH of the base solution can then be found in two ways:found in two ways:– Calculate pOH Calculate pOH
use pH + pOH = 14.00 to determine use pH + pOH = 14.00 to determine pHpH
– Calculate [HCalculate [H++]] use [Huse [H++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14
Then calculate pHThen calculate pH
Step 1: Determine [OHStep 1: Determine [OH--]]
[OH[OH--] = ] = 0.25 mol Ca(OH)0.25 mol Ca(OH)22 x x 2 mol OH2 mol OH--
LL 1 mol Ca(OH)1 mol Ca(OH)22
= 0.50 M= 0.50 M
Strong Bases Strong Bases
Example:Example: Calculate the pH of a 0.25 M Calculate the pH of a 0.25 M Ca(OH)Ca(OH)22 (aq) solution. (aq) solution.
Given: Given: [Ca(OH)[Ca(OH)22] = 0.25 M] = 0.25 M
Find:Find: pH pH
Step 1: Determine [OHStep 1: Determine [OH--]]
[OH[OH--] = ] = 0.25 mol Ca(OH)0.25 mol Ca(OH)22 x x 2 mol OH2 mol OH--
LL 1 mol Ca(OH)1 mol Ca(OH)22
= 0.50 M= 0.50 M
Strong Bases Strong Bases
Step 2: Calculate pOHStep 2: Calculate pOH
pOH = - log [OHpOH = - log [OH--]]pOH = - log (0.50) = 0.30pOH = - log (0.50) = 0.30
Step 3: Calculate pHStep 3: Calculate pH
pH + pOH = 14.00pH + pOH = 14.00
pH = 14.00 - 0.30 = 13.70pH = 14.00 - 0.30 = 13.70
Strong Bases Strong Bases
Example: Example: What is the pH of a solution What is the pH of a solution prepared by mixing 10.0 mL of 0.015 M prepared by mixing 10.0 mL of 0.015 M Ba(OH)Ba(OH)22 and 30.0 mL of 7.5 x 10 and 30.0 mL of 7.5 x 10-3-3 M NaOH? M NaOH?
Given:Given: 10.0 mL of 0.015 M Ba(OH) 10.0 mL of 0.015 M Ba(OH)22
30.0 mL of 7.5 x 1030.0 mL of 7.5 x 10-3 -3 M NaOHM NaOH
Find:Find: pHpH
Strong Bases Strong Bases
Step 1: Find the total [OHStep 1: Find the total [OH--]]
[OH[OH--] = ] = moles OHmoles OH-- from NaOH + mol OH from NaOH + mol OH-- from Ba(OH) from Ba(OH)22
total volume in Ltotal volume in L
= = mmol OHmmol OH-- from NaOH + mmol OH from NaOH + mmol OH-- from Ba(OH) from Ba(OH)22
total volume in mLtotal volume in mL
mmol OHmmol OH--NaOH NaOH == 30.0 mL x 7.5 x 10 30.0 mL x 7.5 x 10-3-3M = M = 0.225 mmol0.225 mmol
mmol OHmmol OH--Ba(OH)2Ba(OH)2 = 10.0 mL x 0.015 M x = 10.0 mL x 0.015 M x 2 mol OH2 mol OH
1 mol Ba(OH)1 mol Ba(OH)22
= = 0.300 mmol0.300 mmol
Strong Bases Strong Bases
Step 1 (cont)Step 1 (cont)
[OH[OH--] = ] = 0.225 mmol + 0.300 mmol0.225 mmol + 0.300 mmol10.0 mL + 30.0 mL10.0 mL + 30.0 mL
= 1.3 x 10= 1.3 x 10-2-2 M M
Step 2: Find pOHStep 2: Find pOH
pOH = - log [OHpOH = - log [OH--] = - log (1.3 x 10] = - log (1.3 x 10-2-2) = 1.89) = 1.89
Strong BasesStrong Bases
Step 3: Find pHStep 3: Find pH
pH + pOH = 14.00pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 1.88 = 12.12pH = 14.00 - pOH = 14.00 - 1.88 = 12.12
4.4 IONIATION CONSTANTS OF 4.4 IONIATION CONSTANTS OF WEAK ACID AND WEAK WEAK ACID AND WEAK BASES (pg 107)BASES (pg 107)
KKaa and K and Kbb as a Measure of Acid and Base as a Measure of Acid and Base
StrenghtStrenght
Weak AcidsWeak Acids
Most acidic substances are Most acidic substances are weak acidsweak acids::– partially ionize in solutionpartially ionize in solution– the solution contains an the solution contains an equilibrium equilibrium
mixturemixture of acid molecules and its of acid molecules and its component ionscomponent ions
CHCH33COCO22HH HH++ (aq) + CH (aq) + CH33COCO22
-- (aq) (aq)
Acetic acidAcetic acid
Weak AcidsWeak Acids
The extent to which a weak acid ionizes The extent to which a weak acid ionizes can be expressed using an equilibrium can be expressed using an equilibrium constant known as the constant known as the acid-dissociation acid-dissociation constant (Kconstant (Kaa).).
For a general reaction:For a general reaction:HX HX (aq) (aq) H H++ (aq)(aq) + X + X- - (aq)(aq)
KKa a = [H= [H++][X][X--]]
[HX][HX]
Note: The rules for Note: The rules for writing an expression for writing an expression for KKaa are the same as those are the same as those
for Kfor Kcc, K, Kp p and K and Kspsp..
Weak AcidsWeak Acids
The magnitude of KThe magnitude of Kaa indicates the indicates the
tendency of the hydrogen ion in an acid tendency of the hydrogen ion in an acid to ionize.to ionize.
– The larger the value of KThe larger the value of Kaa, the stronger , the stronger
the acid is.the acid is.
The pH of a weak acid solution can be The pH of a weak acid solution can be calculated using the initial concentration calculated using the initial concentration of the weak acid and its Kof the weak acid and its Kaa..
Weak AcidsWeak Acids
To calculate the pH of a weak acid To calculate the pH of a weak acid solution:solution:– Write the ionization equilibrium for the Write the ionization equilibrium for the
acid.acid.– Write the equilibrium constant Write the equilibrium constant
expression and its numerical value.expression and its numerical value.– Set up a table showing initial Set up a table showing initial
concentration, change, equilibrium concentration, change, equilibrium concentration.concentration.
– Substitute equilibrium concentrations Substitute equilibrium concentrations into the equilibrium constant into the equilibrium constant expression.expression.
Weak AcidsWeak Acids
To calculate the pH of a weak acid To calculate the pH of a weak acid solution (cont):solution (cont):– Solve for the change in concentration.Solve for the change in concentration.
AssumeAssume that the change in that the change in concentration is small (i.e. < 5%) concentration is small (i.e. < 5%) compared to the initial compared to the initial concentration of the weak acid.concentration of the weak acid.
– Check the validity of previous Check the validity of previous assumptions.assumptions.
– Calculate the final concentrations and Calculate the final concentrations and pH.pH.
Weak AcidsWeak Acids
Example: Example: Calculate the pH of a 0.20 M Calculate the pH of a 0.20 M solution of HCN. Ksolution of HCN. Kaa = 4.9 x 10 = 4.9 x 10-10-10
Step 1: Write the equation for the Step 1: Write the equation for the ionization.ionization.
HCN (aq) HHCN (aq) H++ (aq) + CN (aq) + CN-- (aq) (aq)
Step 2: Write the expression for Ka.Step 2: Write the expression for Ka.
KKaa = [H = [H++] [CN] [CN--] = 4.9 x 10] = 4.9 x 10-10-10
[HCN][HCN]
Weak AcidsWeak Acids
Step 3: Set up a table.Step 3: Set up a table.
HCN (aq) HHCN (aq) H++ (aq) + CN (aq) + CN-- (aq) (aq)
Initial 0.20 M 0.00 M 0.00 M
Change -x +x +x
Equil. 0.20 – x x x
Weak AcidsWeak Acids
Step 4: Substitute equilibrium Step 4: Substitute equilibrium concentrations into the Kconcentrations into the Kaa expression. expression.
KKaa = = (x) (x)(x) (x) = 4.9 x 10 = 4.9 x 10-10-10
(0.20 - x)(0.20 - x)
Step 5: Assume that x << 0.20 M and solve Step 5: Assume that x << 0.20 M and solve for x.for x.
KKaa = = (x) (x)(x) (x) = 4.9 x 10 = 4.9 x 10-10-10
0.200.20xx22 = (0.20) (4.9 x 10 = (0.20) (4.9 x 10-10-10) = 9.8 x 10) = 9.8 x 10-11-11
x = 9.9 x 10x = 9.9 x 10-6-6
Weak AcidsWeak Acids
Step 6: Check the validity of our Step 6: Check the validity of our assumption.assumption.
Is x << 0.20 M Is x << 0.20 M Is x less than 5% of 0.20 MIs x less than 5% of 0.20 M
9.9 x 109.9 x 10-6-6 x 100 = 0.005 % x 100 = 0.005 % 0.200.20
So the assumption is valid.So the assumption is valid.
Weak AcidsWeak Acids
Step 7: Substitute value for x into the Step 7: Substitute value for x into the table to find the [Htable to find the [H++].].
HCN (aq) HHCN (aq) H++ (aq) + CN (aq) + CN-- (aq) (aq)
Initial0.20 M0.00 M0.00 MChange-9.9 x 10-6+9.9 x 10-6+9.9 x 10-6Equil.~ 0.209.9 x 10-69.9 x 10-6
Weak AcidsWeak Acids
Step 8: Calculate the pH using the [HStep 8: Calculate the pH using the [H++]]
pH = - log [HpH = - log [H++] = - log (9.9 x 10] = - log (9.9 x 10-6-6) = 5.00) = 5.00
Weak BasesWeak Bases
Many substances behave as weak bases Many substances behave as weak bases in water.in water.
Weak base + HWeak base + H22O O conjugate acid + OHconjugate acid + OH--
The extent to which a weak base reacts The extent to which a weak base reacts with water to form its conjugate acid and with water to form its conjugate acid and OHOH-- ion can be expressed using an ion can be expressed using an equilibrium constant known as the equilibrium constant known as the base-base-dissociation constant (Kdissociation constant (Kbb).).
Weak BasesWeak Bases
KKbb always refers to the equilibrium in always refers to the equilibrium in which a base reacts with water to form which a base reacts with water to form its conjugate acid and OHits conjugate acid and OH-- ion. ion.
For the reaction:For the reaction:
NHNH33 (aq)(aq) + H + H22O O (l)(l) NH NH44++ (aq)(aq) + OH + OH-- (aq)(aq)
KKbb = [NH = [NH44++] [OH] [OH--]]
[NH[NH33]]Note: The rules for Note: The rules for writing an expression for writing an expression for KKbb are the same as those are the same as those
for Kfor Kcc, K, Kp p and Kand Kspsp..
Weak BaseWeak Base
To calculate the pH of a weak base To calculate the pH of a weak base solution:solution:– Write the ionization equilibrium for the Write the ionization equilibrium for the
base.base.– Write the equilibrium constant Write the equilibrium constant
expression and its numerical value.expression and its numerical value.– Set up a table showing initial Set up a table showing initial
concentration, change, equilibrium concentration, change, equilibrium concentration.concentration.
– Substitute equilibrium concentrations Substitute equilibrium concentrations into the equilibrium constant into the equilibrium constant expression.expression.
Weak Bases Weak Bases
To calculate the pH of a weak base To calculate the pH of a weak base solution (cont):solution (cont):– Solve for the change in concentration.Solve for the change in concentration.
AssumeAssume that the change in that the change in concentration is small (i.e. < 5%) concentration is small (i.e. < 5%) compared to the initial compared to the initial concentration of the weak base.concentration of the weak base.
– Check the validity of previous Check the validity of previous assumptions.assumptions.
– Calculate the [OHCalculate the [OH--] concentration and ] concentration and pOHpOH
– Use pOH to calculate pH.Use pOH to calculate pH.
Weak Bases Weak Bases
Example:Example: Calculate the pH of a 0.20 M solution of Calculate the pH of a 0.20 M solution of methylamine, CHmethylamine, CH33NHNH22. K. Kbb = 3.6 x 10 = 3.6 x 10-4-4..
Step 1: Write the equation for the ionization.Step 1: Write the equation for the ionization.
CHCH33NHNH22 (aq) + H (aq) + H22O CHO CH33NHNH33++ (aq) + OH (aq) + OH-- (aq) (aq)
Step 2: Write the expression for KStep 2: Write the expression for Kbb..
KKbb = [CH = [CH33NHNH33++] [OH] [OH--] = 3.6 x 10] = 3.6 x 10-4-4
[CH[CH33NHNH22]]
Weak BasesWeak Bases
Step 3: Set up a table.Step 3: Set up a table.
CHCH33NHNH2 2 CHCH33NHNH33+ + + OH+ OH--
Initial0.20 M0.00 M0.00 MChange-x+x+xEquil.0.20 – xxx
Weak BasesWeak Bases
Step 4: Substitute equilibrium Step 4: Substitute equilibrium concentrations into the Kconcentrations into the Kbb expression. expression.
KKbb = = (x) (x)(x) (x) = 3.6 x 10 = 3.6 x 10-4-4
(0.20 - x)(0.20 - x)
Step 5: Assume that x << 0.20 M and solve Step 5: Assume that x << 0.20 M and solve for x.for x.
KKbb = = (x) (x)(x) (x) = 3.6 x 10 = 3.6 x 10-4-4
0.200.20xx22 = (0.20) (3.6 x 10 = (0.20) (3.6 x 10-4-4) = 7.2 x 10) = 7.2 x 10-5-5
x = 8.5 x 10x = 8.5 x 10-3-3
Weak BasesWeak Bases
Step 6: Check the validity of our Step 6: Check the validity of our assumption.assumption.
Is x << 0.20 M Is x << 0.20 M Is x less than 5% of 0.20 MIs x less than 5% of 0.20 M
8.5 x 108.5 x 10-3-3 x 100 = 4.2 % x 100 = 4.2 % 0.200.20
So the assumption is valid.So the assumption is valid.
Weak BasesWeak Bases
Step 7: Substitute value for x into the Step 7: Substitute value for x into the table to find the [OHtable to find the [OH--].].
CHCH33NHNH2 2 CHCH33NHNH33+ + + OH+ OH--
Initial0.20 M0.00 M0.00 MChange-8.5 x 10-3+8.5 x 10-3+8.5 x 10-3Equil.~ 0.208.5 x 10-38.5 x 10-3
Weak AcidsWeak Acids
Step 8: Calculate the pOHStep 8: Calculate the pOH
pOH = - log [OHpOH = - log [OH--] = - log (8.5 x 10] = - log (8.5 x 10-3-3) = 2.07) = 2.07
Step 9: Calculate the pHStep 9: Calculate the pH
pH + pOH = 14.00pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 2.07 = 11.93pH = 14.00 - pOH = 14.00 - 2.07 = 11.93